Imagine a square made up of 1x1 squares, of which for a square of length n, there are n² of these squares..

This is our "garden".

Now.. one of these squares has to be for a shrub..

We're given 3 different types of paving slabs.. first of an L shaped one (e.g. 3 squares, cant really explain other than that), then a 1x3 and finally a 1x4.. we have to work out for what n these can work (having started off with the L's.. the "company" then suggests the next options)

The shrub I must add has to be able to be present in any of the 1x1 squares..

Now, we've proved what n will definitely not work using the following ideas:

The square of side n must fit the pattern (n²-1)/3 ∈ Z+ Now, n can be written as or n=3m+p, where p=0,1 or 2 and m=0,1,2 Thus, n² = 9m² +6mp + p²And, (n²-1)/3 = (9m² +6mp + p² -1)/3 = 3m² +2mp + (p² -1)/3The co-efficients for both m² and mp are divisible by 3 so thus whether the whole quadratic is divisible by 3 is dependant on this p² -1 term.Given the values of p earlier for any n, p²=0,1 or 4. And obviously (p² -1)/3 therefore equals either -⅓, 0 or 1. Thus the pattern will only work for squares that are not divisible by 3.

This would work for both L and 1x3 logically as they are both composed of 3 squares.. and a similiar idea would be used for 1x4..

The thing we want to know is that how can we prove that for any n (other than those discounted above) they will work for the L-shaped blocks (which seemingly we've found simply by trial)..

Also the 1x3 and 1x4 are both worse of in terms of their use.. but how can we show this mathematically.. i presume between the Ls and 1x3s that it's due to the fact that the L has a rotational symmetry order of 1 as opposed to the 1x3s order of 2.. thus meaning it can create more shapes when put with others..but again this is all worded and not particuarly mathematical...

we have a presentation on this, first year group work basically (designed no doubt in a way to build team skills..) we'd noticed a lot of the patterns between us however were unsure what to make of them or how to prove them :s

well the point being is that it's the sum to infinity, with the terms ever approaching x² (with use of the limit, which is a standard point) then each f(x) is approaching x² and thus it's a divergant series adding up to a sum that approaches infinity as x->infinity.

The value of x is the point on the x-axis that the quadratic function touches it, if it touches once then there's one solution however it can cross it then come back up as it's a curve and thus there are 2 solutions.

Now, you know there's only one solution at (3,0).. e.g. x=3.

therefore the family logically is:

a(3²)+b(3)+c=0

therefore:

9a+3b+c=0

(or alternatively as c is whats called a dummy variable, you could simply write it using a new value and simplify it)

looking at it again.. (and again, etc) i'm thinking it suggests the number of parcels is the number of combinations of dimensions (a,b,c) with the bound that is set as n.. I'm going to email him now to see if that's correct.. any suggestions if that is the case?

Hey,we've been set an assignment in pure maths and have no clue how to go about it..

"A student on a year abroad buys a number of presents and wants to send them home. The goods are already wrapped up as individual packets: by a curious chance there is an integer n such that there is precisely one packet of depth a, width b and length c, for each set of integers (a,b,c) with 1<=a<=b<=c<=n

The student could send each packet home separately, but it is cheaper (and mathematically more interesting) to make the packets up into parcels. To avoid breakages, the parcels must have no empty spaces (the country is experiencing a shortage of bubble wrap). In other words, a parcel of size pxqxr will have volume pqr that is equal to the sum of the packets it contains. Assuming that it does not matter how large the parcels are, how many parcels does the student need?"